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Kinetic Theory
Physics · Class 11NCERT pp. 244–2562 sets × 25 questions
Key facts and formulas for Kinetic Theory, three worked sample questions, and two full practice sets. Every question has a hint, a reason for every option and a step-by-step solution, with a calculator or animation where one helps. English and Hinglish.
Formula sheet for this chapter
| Idea | Rule or fact |
|---|
| Ideal gas law | PV = μRT = N k_B T, so n = P/(k_B T). R = 8.314 J mol⁻¹ K⁻¹, k_B = 1.38 × 10⁻²³ J K⁻¹, N_A = 6.02 × 10²³ |
| Moles and molecules | μ = mass / M₀, N = μ N_A, mass of one molecule = M₀ / N_A. At STP one mole = 22.4 L |
| Dalton's law | Partial pressure = mole fraction × total pressure. Total pressure = sum of partial pressures |
| Kinetic pressure | P = ⅓ n m v², and PV = (2/3) E for total translational kinetic energy E |
| Mean kinetic energy | ½ m v² = (3/2) k_B T, depends on temperature only. Total E = (3/2) μ R T |
| Speeds | v_mp = √(2RT/M₀) < v_avg = √(8RT/πM₀) < v_rms = √(3RT/M₀). Use M₀ in kg mol⁻¹ |
| Graham's law | Rate of diffusion or effusion ∝ 1/√M; time for a fixed volume ∝ √M |
| Mean free path | l = 1/(√2 n π d²), so l ∝ 1/n. n = P/(k_B T), so l doubles if P halves at fixed T |
| Degrees of freedom | C_v = (f/2)R, C_p = C_v + R, γ = 1 + 2/f. Monatomic f = 3, rigid diatomic f = 5, nonlinear f = 6 |
| Mixtures and solids | Mixture: average C_v and C_p first, then γ = C_p/C_v. Solid: C = 3R, specific heat s = C/M₀ |
Three sample questions
Q1. Statement I: Between collisions, a molecule of a gas moves in a straight line with constant velocity.
Statement II: Interatomic forces are of very short range, so they are negligible when molecules are far apart.
Choose the correct option.
- Both are true but II does not explain I
- Both are true and II explains I
- I is true and II is false
- I is false and II is true
Hint and solution
Hint: Ask whether statement II gives the reason why no force acts on the molecule between collisions. Think of Newton's first law.
- Between collisions: net force = 0 — Check statement I. In a dilute gas a molecule feels no force between collisions, so by Newton's first law it moves in a straight line with constant velocity. Statement I is true.
- Forces act only at ~ 1 Å range — Check statement II. Interatomic forces are of very short range, so they matter only when molecules almost touch. Statement II is true.
- II is the reason for I — Molecules are far apart compared with the force range, so the force is negligible between collisions. That is exactly the reason for straight line motion.
- Both true, II explains I — So both statements are true and II explains I. Option B.
Answer: B. In a gas the spacing is large, so forces are negligible between collisions and Newton's first law applies.
Q2. Match the person with the idea.
I. Kanada II. Democritus III. Dalton IV. Avogadro
P. Equal volumes of gases hold equal numbers of molecules Q. Paramanu of four kinds R. Atoms combine in small whole-number ratios S. Smooth round atoms flow, rough atoms lock
- I–S, II–Q, III–R, IV–P
- I–Q, II–R, III–S, IV–P
- I–P, II–S, III–R, IV–Q
- I–Q, II–S, III–R, IV–P
Hint and solution
Hint: Kanada is an ancient Indian thinker whose word for an atom was paramanu. Democritus explained solids and liquids by atom shapes.
- Dalton → whole-number ratios (R) — Dalton's atomic theory says atoms combine in small whole-number ratios. So III matches R.
- Avogadro → equal volumes (P) — Avogadro's law says equal volumes of gases at the same temperature and pressure hold equal numbers of molecules. So IV matches P.
- Kanada → Q, Democritus → S — Kanada described paramanu of four kinds. Democritus said smooth round atoms flow while rough ones lock. So I-Q and II-S.
- I–Q, II–S, III–R, IV–P — Only one option has all four pairs correct. The answer is I-Q, II-S, III-R, IV-P. Option D.
Answer: D. Kanada: paramanu. Democritus: shapes of atoms. Dalton: whole-number ratios. Avogadro: equal volumes.
Q3. For air at STP the mean free path is of the order of
- 10⁻⁷ m
- 10⁻¹⁰ m
- 10⁻³ m
- 10⁻¹ m
Hint and solution
Hint: The mean free path is roughly a thousand times the molecular diameter, and a molecule is about 2 angstrom wide.
- l = 1 / (√2 n π d²) — The mean free path is the average distance between two collisions. It depends on the number density n and molecular diameter d.
- n ≈ 2.7 × 10²⁵ m⁻³, d ≈ 2 × 10⁻¹⁰ m — At STP the number density of air is about 2.7 × 10²⁵ per cubic metre and a molecule is about 2 angstrom wide.
- l ≈ 2 × 10⁻⁷ m ≈ 1000 d — Putting the values in gives about 2 × 10⁻⁷ metre, roughly a thousand molecular diameters.
- Order is 10⁻⁷ m — So the order of magnitude is 10⁻⁷ m. Option A.
Answer: A. Calculation with d ≈ 2 Å gives about 2 × 10⁻⁷ m, roughly a thousand molecular diameters.
How to use this chapter
Read the NCERT pages first, then attempt Set 1 without the hints. Note every wrong answer in a mistake list, read the reason for each option, and retry Set 2 two days later. A topic is solid when you can say why each wrong option is wrong.